Question

The amount of money in a savings account over time is given in the table below. Answer the questions below to determine what kind of function would best fit the data, linear or exponential.
Number of Years Since Account Opened, x 11 22 33 44
Money in account ($), f(x) 230, point, 3, 0230.30 491, point, 6, 0491.60 750, point, 8, 2750.82 1, comma, 009, point, 5, 21,009.52
(a linear or an exponential) function would better model the data because as x increases, the y values change (additively or multiplicatively) The (common difference/slope or common ratio/multiplier/base) of this function is approximately (???)

Answer 1

To determine whether a linear or exponential function best fits the data in the given table, we need to analyze how the money in the account changes as the number of years since the account was opened increases.

### Given Data:
- **Years (x)**: 11, 22, 33, 44
- **Money (f(x))**:
- f(11) = $230.30
- f(22) = $491.60
- f(33) = $750.82
- f(44) = $1,009.52

### Step 1: Calculate the changes in money to determine if the changes are additive or multiplicative.

- From 11 to 22 years:
- Change = 491.60 - 230.30 = 261.30
- From 22 to 33 years:
- Change = 750.82 - 491.60 = 259.22
- From 33 to 44 years:
- Change = 1,009.52 - 750.82 = 258.70

### Step 2: Check if the changes are additive (linear) or multiplicative (exponential).

1. **Additive Changes**: The differences between the amounts are relatively consistent but slightly decreasing:
- 261.30, 259.22, 258.70

2. **Percentage Changes**:
- From 230.30 to 491.60:
- Percentage Increase = (491.60 - 230.30) / 230.30 * 100 ≈ 113.51%
- From 491.60 to 750.82:
- Percentage Increase = (750.82 - 491.60) / 491.60 * 100 ≈ 52.71%
- From 750.82 to 1,009.52:
- Percentage Increase = (1,009.52 - 750.82) / 750.82 * 100 ≈ 34.41%

The percentage increases are decreasing, suggesting that the function is not perfectly exponential, but the increase in absolute values suggests a strong possibility of exponential growth as opposed to linear growth.

### Conclusion:
Based on the analysis, an **exponential function** would better model the data because as \( x \) increases, the \( y \) values change **multiplicatively**. The **common ratio/multiplier** is calculated using the changes in the amounts:

Here’s a rough estimate of the ratio using the first two values:
- Ratio = \( \frac{f(22)}{f(11)} \) ≈ \( \frac{491.60}{230.30} \) ≈ 2.14

However, to find a general multiplier, you would ideally use regression or further calculations across all values but based on this rough estimate, an approximate **multiplier** could be around **2.14**.

Thus, the exponential function would better fit this situation with the approximate multiplier being around 2.14.